# Math 210B:Algebra, Homework 2

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1 Math 210B:Algebra, Homework 2 Ian Coley January 21, 2014 Problem 1. Is f = 2X 5 6X + 6 irreducible in Z[X], (S 1 Z)[X], for S = {2 n, n 0}, Q[X], R[X], C[X]? To begin, note that 2 divides all coefficients of f. Since 2 Z is not a unit, f is reducible in Z[X]. However, in all other cases, 2 is a unit, so we do not have this trivial factorisation. Consider the polynomial f = X 5 3X + 3 = f/2. This polynomial is irreducible in Z[X] by Eisenstein s criterion. Further, it is monic, hence primitive, so f is irreducible in Q[X] as well. This also implies that f is irreducible in (S 1 Z)[X], since this polynomial ring is a subring of Q[X]. Evidently since f and f differ only by a unit in Q[X] and (S 1 Z)[X], they are both irreducible or both reducible. Hence f is irreducible in these two polynomial rings. We now consider the graph of f in R[X]. At a = 1, we have f(a) = 2 > 0. At b = 2, we have f(b) = 46 < 0. Therefore by the intermediate value theorem (which we may apply since polynomial functions are continuous), there must be some c [a, b] such that f(c) = 0. Therefore f has a root in R[X]. This yields a factorisation f = (X c)g, where g = d i X i is a degree four polynomial. Suppose that g / R[X]. Then some d i C \ R. It is clear that we do not have i = 4 or i = 0. Then multiplying, the term in f X i (d i 1 + c d i ) must have a real coefficient. Since c d i C \ R, we must have d i 1 C \ R as well. Repeating this process, we see that we must have d 0 C \ R, which is a contradiction. Therefore f = (X c)g is a factorisation in R[X], so f is reducible. By the Fundamental Theorem of Algebra, f has a root in C. Let α 1 be this root. Then we get a reduction f = (X α)g, where g is a degree four complex polynomial. Repeating this process, we obtain a factorisation f = 5 (X α i ), α i C. i=1 Hence f is reducible in C[X]. Problem 2. 1

2 (a) A multiplicative subset S in a commutative ring R is called saturated if whenever x s for x R and s R, then x S. For any multiplicative subset S R, define the saturation S of S: S = {x R : x s for some s S. Show that S is a saturated multiplicative subset. (b) Let S R be a multiplicative subset. Show there is a natural ring isomorphism S 1 R to S 1 R. (a) First, we show that S is multiplicative. Suppose x, y S, and let s, t S such that x s and y t. Then xy st S, so xy S as required. Now suppose x R and s S such that x s. Then since s S, there exists some t S such that s t. But then x t as well, so we have x S. Therefore S is saturated. (b) Let ϕ : S 1 R S 1 R be the inclusion map, which is sensible since S S. Then we claim that ϕ is an isomorphism. That is is a ring homomorphism is clear by construction. Now, suppose that ϕ(r/s) = 0. Then since we may write 0/s for the equivalence class corresponding to the zero element, we must have for some t S, t(rs 0r) = trs = 0. Since t S, we know that t z for z S. Write z = yt. Then zrs = y(trs) = y(0) = 0. Hence r/s = 0 in S 1 R as well, so ϕ is injective. Now take any element r/s S 1 R. Then s z for some z S. Write z = ys. Then ( yr ) ϕ = yr z ys = r because (yrs rys) = 0. s Therefore ϕ is surjective as well. Hence ϕ is an isomorphism under the natural inclusion map, and we are done. Problem 3. Let S be a multiplicative subset of a commutative ring R. For any R-module M, define the localisation S 1 M as a module over S 1 R. Show that the correspondence M S 1 M extends to a functor R-Mod S 1 R-Mod. Recall that we defined the localisation S 1 R by considering formal pairs (s, r) S R modulo an equivalence relation. In the same way, we let S M be an R-module in the following way: r (t, m) = (st, r m) s 2

3 where r m derives from the R-module structure of M. Further, we identify elements of S 1 M under the equivalence relation (s, m) (t, n) u S such that u(sn tm) = 0. Therefore we have related to each R-module M a S 1 R-module S 1 M. Therefore let F : R-Mod S 1 R-Mod be the functor where F (M) = S 1 M. For the action on morphisms, suppose that ϕ : M N is an R-module homomorphism. Then we let F (ϕ) : S 1 M S 1 N such that F (ϕ)(m/s) = ϕ(m)/s. As a result, tϕ(m) + sϕ(n) F (ϕ)(m/s + n/t) = = F (ϕ)(m/s) + F (ϕ)(n/t) st F (ϕ)(r/t m/s) = rϕ(m) = (r/t) F (ϕ)(m/s). ts Therefore F (ϕ) is an S 1 R-module homomorphism, so F is in fact a functor. Problem 4. Let I be a (two-sided) ideal of a ring R and let M be a (left) R-module such that IM = 0. Show that M has a natural structure of a (left) R/I-module. We define the structure thus: let r + I R/I. Then for m M, we let (r + I) m = r m. Suppose that r + I and r + I were two representatives of the same coset. Then we may write r = r + a for a I. Hence (r + I) m = (r + a + I) m = (r + a) m = r m + a m = r m = (r + I) m since IM = 0. Therefore this action is well defined. That the action is appropriately distributive follows from the R-module structure of M. Finally, since (r + I)(s + I) = rs + I, we have (rs + I) m = rs m = r (s m) = (r + I) ((s + I) m) as required. Therefore our R/I-module structure is sound. Problem 5. Let {M i }, N be (left) R-modules. Show that there are natural isomorphisms ( ) ( Hom R M α, N = Hom R (M α, N) and Hom R N, ) M α = Hom R (N, M α ) Let p β : M α M β be projection onto the βth component and let i β : M β M α be inclusion. Let f Hom R ( M α, N). Then f i β is a homomorphism from M β to N for each β A. Therefore we can associate to each such f the family of homomorphisms f i α. 3

4 Conversely, let f α Hom R (M α, N), where each f α : M α N. Then f β p β is a homomorphism from M α to N for each β A. Taking all of these together, we let f : M α N by ( ) f mα = (f ( ) α p α ) mα = f α (m α ). This sum is well-defined since coproducts are nonzero in only finitely many places. These maps are mutually inverse. As in the first part, consider f and f i α. Then we have the induced homomorphism f where ( ) f mα = (f ( ) i α p α ) mα = ( ) f(m α ) = f mα. Similarly, given f α and f from the second part, there is an induced homomorphism f α. We see f α = f i α = (f α p α ) i α = f α. So this is inverse as well. Therefore this association is one-to-one, so the two sets of homomorphisms are isomorphic. Now, let f Hom (N, M α ). Then there is a homomorphism f α = p α f from N to M α. Hence f α Hom(N, M α ). Similarly, let f α Hom(N, M α ). Then f = i α f α is a map from N to M α. These are immediately seen to be mutually inverse, so the two sets are isomorphic. Problem 6. Prove that a (left) R-module over a ring R with identity generated by one element is isomorphic to R/I, where I is a (left) ideal in R. Let M be that module, and let x M be its generator. Then we claim that M = R/I as modules, where I is the annihilator of x, namely I = {r R : r x = 0}. First, I is a left ideal, because for any s R and r I, sr x = s (r x) = s 0 = 0. Second, we define an R-module homomorphism ϕ : R M, ϕ(r) = r x Then ϕ is surjective, since M = Rx. Further, the kernel of ϕ is precisely {r R : r x = 0} = I. Therefore by the first isomorphism theorem, R/I = M. This completes the proof. Problem 7. Let F (A) = A tor be the functor from the category of abelian groups to itself, where A tor is the subgroup of elements of finite order in A. Show that F is left exact. 4

5 Let 0 A f B g C 0 be an exact sequence of abelian groups. Then we would like to show that 0 A tor f Btor ḡ Ctor is also exact, where F (f) = f, and similarly the others. Suppose ϕ : G H is a group homomorphism, and g G tor. Then n g = 0 for some n N. Hence 0 H = ϕ(0 G ) = ϕ(n g) = n ϕ(g), so ϕ(g) H tor. Hence in our case, we have f : A B is an injective group homomorphism. Then f : A tor B tor is also injective, since if f(a) = f(a ), then f(a) = f(a ), which would be a contradiction. Now we only need to show im f = ker ḡ. It is clear that im f = B tor im f and ker ḡ = B tor ker g. Therefore since ker g = im f, intersecting with B tor preserves equality. Hence im f = ker ḡ, so the sequence is exact. Therefore the functor is left exact. Problem 8. Let a b c d M 1 M 2 M 3 M 4 M 5 f 1 f 2 f 3 f 4 f 5 N 1 a N 2 b N 3 c N 4 d N 5 be a commutative diagram of (left) R-modules and R-homomorphisms with exact rows. Prove that (a) if f 1 is surjective, f 2, f 4 are injective, then f 3 is injective; (b) if f 5 is injective, f 2, f 4 are surjective, then f 3 is surjective. (a) Suppose that f 3 (m) = 0. Then c (f 3 (m)) = 0 N 4 as well. Then by the diagram, f 4 (c(m)) = 0, and since f 4 is injective, c(m) = 0. By exactness of the top row, ker c = im b, so there exists n M 2 such that b(n) = m. Since we have f 3 (b(n)) = 0, by the diagram chase we have b (f 2 (n)) = 0. Since ker b = im a, there exists some p N 1 such that a (p) = f 2 (n). Since f 1 is surjective, there exists q M 1 such that f 1 (q) = p. Since f 2 (n) = a (f 1 (q)) = f 2 (a(q)) and f 2 is injective, a(q) = n. Therefore m = b(a(n)). Since b(a(n)) = 0 by exactness, m = 0. Since f 3 (m) = 0 implies m = 0, f 3 is injective. (b) Let n N 3. Then since f 4 is surjective, c (n) has a nontrivial preimage, so c (n) = f 4 (m) for m M 4. Then by commutativity, d (f 4 (m)) = f 5 (d(m)). Since im c = ker d, 5

6 d (f 4 (m)) = d (c (n)) = 0. Since f 5 (d(m)) = 0 and f 5 is injective, d(m) = 0, so m ker d = im c. Therefore there exists l M 3 such that c(l) = m. Then c (n) = f 4 (m) = f 4 (c(l)) = c (f 3 (l)), so n f 3 (l) ker c = im b. Therefore let p N 2 such that b (p) = n f 3 (l). Since f 2 is surjective, there exists q in M 2 so that f 2 (q) = p. By commutativity, f 3 (b(q)) = c (f 2 (q)) = n f 3 (l). Therefore f 3 (l + b(q)) = n. Therefore f 3 is surjective, so we are done. Problem 9. Let M i be (left) R i -modules for i = 1, 2,..., n. Show that M = M 1 M n has a natural structure of a (left) module over the ring R = R 1 R n. Prove that any (left) R-module is isomorphic to M as above for some (left) R i -modules M i. M is an abelian group under the usual product of abelian groups. The natural structure of M as an R-module is (r 1,..., r n ) (m 1,..., m n ) = (r 1 m 1,..., r n m n ). Now let M be any R-module. Then consider the projections p i : M M i by p i (m) = (0,..., 1 Ri,..., 0) m where 1 is in the ith spot. Call that coefficient e i R. Then we claim that M is a direct product of these M i. Indeed, each M i is an R i module. Further, each m M has a unique expression as e i m, corresponding to (e 1 m,..., e n m) n i=1 M i. Therefore we are done. Problem 10. Let R = Z[X, Y ]. Construct an exact sequence of R-modules 0 R R R R f Z 0 where f(g(x, Y )) = g(0, 0). Here Z is viewed as an R-module via X 1 = 0 = Y 1. We will give a sequence 0 R ψ R R ϕ R f Z 0. Working backwards, we see that f must be surjective. Now we consider ker f. Since f(g(x, Y )) = g(0, 0), we know ker f = {g R : g has no constant term}. Therefore this must be the image of ϕ : R R R. Let ϕ((g(x, Y ) h(x, Y )) = X g+y h. Then ker f im ϕ. We claim the reverse inclusion is also true. Consider any polynomial g(x, Y ) = a ij X i Y j n i,j 0 6

7 with no constant term, i.e. a 00 = 0. Then for all terms of the form a i0 X i, consider the polynomial g X = a i0 X i 1. For all other terms a ij X i Y j, consider the polynomial g Y = aij X i Y j i. Then ϕ(g X g Y ) = X g X + Y g Y = a i0 X i + j>0 a ij X i Y j = g. Therefore g im ϕ, so ker f = im ϕ. We now examine ker ϕ. We have ϕ(g h) = 0 if and only if X g = Y h, so we define ψ : R R R by ψ(g(x, Y )) = Y g X g. Then by construction, im ψ ker ϕ. Conversely, suppose that g h ker ϕ. Then since X g = Y h, we must have Y g and X h. Therefore let g = Y g and h = X h. Then X g = XY g = XY h = Y h. Since we may cancel terms in a domain, we have g = h. Hence ψ(g ) = Y g X g = g h, so im ψ = ker ϕ. Now examining the first terms of the sequence, we need ker ψ = 0. Since ψ is injective by construction, this completes the proof. 7

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